The sum of the 4th and 6th term of an AP is 42 the sum of the 3rd and 9th term of the progression is 52. find the first term, the common difference and the 20th term of the progression
Let the first term be $a$ and the common difference be $d$. Then, we know that:
$4^{th}$ term $= a + 3d$
$6^{th}$ term $= a + 5d$
$3^{rd}$ term $= a + 2d$
$9^{th}$ term $= a + 8d$
We can use the information given in the problem to set up the following equations:
$a+3d + a+5d = 42$ (sum of $4^{th}$ and $6^{th}$ terms)
$a+2d + a+8d = 52$ (sum of $3^{rd}$ and $9^{th}$ terms)
Simplifying these equations gives:
$2a+8d = 42$
$2a+10d = 52$
Subtracting the first equation from the second gives:
$2d = 10$
So, $d = 5$. Substituting this value of $d$ into either of the two equations above gives:
$a = 11$
Therefore, the first term is $11$ and the common difference is $5$. To find the $20^{th}$ term, we can use the formula for the $n^{th}$ term of an arithmetic progression:
$t_n = a + (n-1)d$
Substituting $n=20$, $a=11$, and $d=5$ gives:
$t_{20} = 11 + (20-1)5 = 106$
Therefore, the $20^{th}$ term of the progression is $106$.
Thanks
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a town x is 400m SW to town y what is the ground distance between town x and y
We can use the Pythagorean theorem to find the ground distance between the towns. Let's assume that town X is located at the origin (0,0) and town Y is located at point (a,b) where a and b are the distances along the east-west and north-south axes respectively. Since we know that town X is located 400m to the southwest of town Y, we can conclude that a and b are related as follows:
a = b + 400
Now, let's find the distance between the two points using the Pythagorean theorem:
$
\begin{aligned}
\text{Distance} &= \sqrt{(a-0)^2 + (b-0)^2} \\
&= \sqrt{a^2 + b^2} \\
&= \sqrt{(b+400)^2 + b^2} \\
&= \sqrt{2b^2 + 800b + 160000} \\
&= \sqrt{2(b^2 + 400b + 80000)} \\
\end{aligned}
$
We can complete the square inside the square root by adding and subtracting $(400)^2$, which gives:
$
\begin{aligned}
\text{Distance} &= \sqrt{2(b^2 + 400b + 400^2 - 400^2 + 80000)} \\
&= \sqrt{2((b+200)^2 - 36000)} \\
&= \sqrt{2(b+200)^2 - 2(600)^2} \\
&= \sqrt{2(b+200+600)(b+200-600)} \\
&= \sqrt{2(b+800)(b-400)} \\
\end{aligned}
$
Therefore, the ground distance between town X and town Y is $\sqrt{2(b+800)(b-400)}$ meters, where $b$ is the distance along the north-south axis from town X to town Y.
from a place 400m north of X, a student walks eastwards to a place Y which 800m from X. What is the bearing of X from Y
To find the bearing of X from Y, we need to determine the angle between the line XY and the north direction. Let's first draw a diagram:
```
Y
/|
/ |
/ |
/θ | 800m
/ |
/_____|_______ X
400m
```
We can see that the angle θ is opposite the side that is 400m long. Moreover, the length of XY is the hypotenuse of a right-angled triangle with sides 400m and 800m. Therefore, we can use the Pythagorean theorem to find the length of XY:
$
\begin{aligned}
XY^2 &= 400^2 + 800^2 \\
&= 640000 \\
\end{aligned}
$
Taking the square root of both sides gives:
$XY = \sqrt{640000} = 800\sqrt{2}$ meters
Now, we can use trigonometry to find the value of θ:
$
\begin{aligned}
\tan(\theta) &= \frac{400}{800\sqrt{2}} \\
&= \frac{1}{2\sqrt{2}} \\
\end{aligned}
$
Taking the inverse tangent of both sides gives:
$\theta = \tan^{-1}\left(\frac{1}{2\sqrt{2}}\right) \approx 19.5^\circ$
Therefore, the bearing of X from Y is approximately $\boxed{019.5^\circ}$ (measured clockwise from north).